Question: Casey ran out of time while taking a multiple-choice test and plans to guess on the last $10$ questions. Each question has $5$ possible choices, one of which is correct. Let $X=$ the number of answers Casey correctly guesses in the last $10$ questions. Which of the following would find $P(X=2)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\left( \dfrac15 \right)^2\left( \dfrac45 \right)^8$ (Choice B) B ${10 \choose 5} \left( \dfrac15 \right)^2\left( \dfrac45 \right)^8$ (Choice C) C ${10 \choose 5} \left( \dfrac15 \right)^8\left( \dfrac45 \right)^2$ (Choice D) D ${10 \choose 2} \left( \dfrac15 \right)^2\left( \dfrac45 \right)^8$ (Choice E) E ${10 \choose 2} \left( \dfrac15 \right)^8\left( \dfrac45 \right)^2$
Solution: Probability of $2$ successes We want the probability that there are $2$ successes (correct answers) in $10$ trials (number of questions), so we're going to need $8$ failures (wrong answers) as well. The probability of each success is ${\dfrac15}$ and the probability of each failure is $\dfrac45}$. We can multiply probabilities to find the probability of getting $2$ successes followed by $8$ failures: $\begin{aligned} P(\text{SSFFFFFFFF})&=\left({\dfrac15}\right)\left({\dfrac15}\right)\left(\dfrac45}\right)\left(\dfrac45}\right)\dots\left(\dfrac45}\right) \\\\ &=\left({\dfrac15}\right)^2\left(\dfrac45}\right)^8 \end{aligned}$ The binomial coefficient ${n \choose k}$ SSFFFFFFFF isn't the only arrangement that produces $2$ successes in $10$ trials. For instance, FFFFFFFFSS would also produce the desired outcome. To count how many possible arrangements there are, we use the binomial coefficient ${n \choose k}$. It tells us the number of possible arrangements for $k$ successes in $n$ trials. In this problem, we want $k=2$ successes (correct answers) in $n=10$ trials (questions), so we should use the binomial coefficient ${10 \choose 2}$. [Tell me more about the binomial coefficient.] Putting it together Each arrangement has probability $\left( \dfrac15 \right)^2\left( \dfrac45 \right)^8$ so for our final answer we multiply this probability by the number of possible arrangements: ${10 \choose 2} \left( \dfrac15 \right)^2\left( \dfrac45 \right)^8$ The answer: ${10 \choose 2} \left( \dfrac15 \right)^2\left( \dfrac45 \right)^8$